What is the domain and range of #y=-sqrt(x^2-3x-10)#?

Question

Eli Assouline · Accepted Answer

Domain: #(-oo, -2] uu [5, + oo)#
Range: #(-oo, 0]# Since you can only take the square root of positive numbers for real numbers, you can tell from the outset that the expression under the square root must be positive. To find the square root's roots, set the expression inside it to zero. #x^2 - 3x - 10 = 0# #x_(1,2) = (-(-3) + - sqrt((-3)^2 - 4 * 1 * (-10)))/(2 * 1)# #x_(1,2) = (3 +- sqrt(49))/2# #x_(1,2) = (3 +- 7)/2 = {(x_1 = (3 + 7)/2 = 5), (x_2 = (3 - 7)/2 = -2) :}# The two roots of the quadratic can be used to factor it. #(x-5)(x+2) = 0# In order for this to be greater than or equal to zero, you need either both #(x-5)# and #(x+2)# to be positive or both to be negative. Do not forget to include the values that makes each term equal to zero. For any value of #x>=5# you have #{(x-5>=0), (x+2 > 0) :} implies (x-5)(x+2)>=0# Likewise, for any value of #x <= -2# you have #{(x-5<0), (x+2 <= 0) :} implies (x-5)(x+2)>=0# The domain of the original function will thus be #(-oo, -2] uu [5, + oo)#. To find the range of the function, keep in mind that the square root of any positive real number is always positive. Since you have #f(x) = - sqrt(x^2 - 3x - 10)# you get that #f(x) <= 0", "(AA)x in (-oo, -2] uu [5, + oo)# The range of the function will thus be #(-oo, 0]#. graph{-sqrt(x^2 -3x - 10) [-10, 10, -5, 5]}

Genesis Allen · Answer

undefined The domain of $y = -\sqrt{x^2-3x-10}$ is $x \leq -2$ or $x \geq 5$, and the range is $-\infty < y \leq 0$.

Nathan Axel · Answer

undefined Domain: All real numbers such that $ x^2 - 3x - 10 \geq 0 $ (ensuring the square root is real).
Range: All real numbers less than or equal to 0.